Question

Given is the graph of the function h(x)=log12x
. Which of the following correctly uses the graph to find h(1)+h−1(−1)
?

(1 point)
Responses

h(1)+h−1(−1)=0
h left parenthesis 1 right parenthesis plus h superscript negative 1 baseline left parenthesis negative 1 right parenthesis equals 0

h(1)+h−1(−1)=2
h left parenthesis 1 right parenthesis plus h superscript negative 1 baseline left parenthesis negative 1 right parenthesis equals 2

h(1)+h−1(−1)=3
h left parenthesis 1 right parenthesis plus h superscript negative 1 baseline left parenthesis negative 1 right parenthesis equals 3

h(1)+h−1(−1)=2.5

Answer 1

To find \( h(1) + h^{-1}(-1) \) using the graph of the function \( h(x) = \log_{12}(x) \), we need to:

1. **Calculate \( h(1) \)**:
\[
h(1) = \log_{12}(1)
\]
Since \( \log_a(1) = 0 \) for any base \( a \), we have:
\[
h(1) = 0
\]

2. **Find \( h^{-1}(-1) \)**:
- \( h(x) = y \) can be written in the inverse form as \( x = h^{-1}(y) \).
- We need to solve \( h(x) = -1 \):
\[
\log_{12}(x) = -1
\]
This means:
\[
x = 12^{-1} = \frac{1}{12}
\]
Therefore, \( h^{-1}(-1) = \frac{1}{12} \).

3. **Combine the results**:
We found that:
\[
h(1) = 0
\]
\[
h^{-1}(-1) = \frac{1}{12}
\]
Now, we add these together:
\[
h(1) + h^{-1}(-1) = 0 + \frac{1}{12} = \frac{1}{12}
\]

Since none of the provided options directly state \( \frac{1}{12} \) as a possible answer, we can surmise that perhaps there is misunderstanding with the negative one; it is crucial to confirm by looking at the graph for \( h \) and its inverse for the behavior in the specified values. However, based on our calculations for given \( h(1) + h^{-1}(-1) \), none of the provided options are correct.

Thus, reviewing provided options:

- \( h(1) + h^{-1}(-1) = 0 \) → Incorrect
- \( h(1) + h^{-1}(-1) = 2 \) → Incorrect
- \( h(1) + h^{-1}(-1) = 3 \) → Incorrect
- \( h(1) + h^{-1}(-1) = 2.5 \) → Incorrect

Correct calculation yields \( \frac{1}{12} \).